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The differential equation an equation containing the derivatives of one or more dependent variable with respect to one or more independent variable . and the differential equation stand out dramatically in Physics , Biology, and chemistry … A chemical changes lead to the formation of substances that help grow our food , make our lives more productive , a chemical change or chemical reaction is a process in which one or more pure substances are converted into one or more different pure substances , The Chemical changes take place as one substance is converted into another , it Can the conditions be altered to speed the changes up, slow them down, or perhaps reverse them , Once chemists understand the nature of one chemical change, they begin to explore the possibilities that arise from causing similar changes Chemical*…show more content…*

The resulting reaction between the two chemicals is such that for each gram of A, 4 grams of B is used. It is observed that 30 grams of the compound C is formed in 10 minutes Determine the amount of C at time t if the rate of the reaction is proportional to the amounts of A and B remaining and if initially there are 50 grams of A and 32 grams of B. How much of the compound C is present at 15 minutes? Interpret the solution as t →∞ ?

In this example we will use the rate equation .

The rate equation: is the rate which a chemical A transforms into a second chemical B is proportional to the amount Q of A remaining untransformed at time t ,

DE= dQ/dt=KQ , K>0

Data:

A + 4B → C x(t):g X(0) = 0 , x(10)=30 , a=50g , b=32 , M=1 , N=4 , x(15)=?

Solution:

1) a - M/(M+N ) X >>> 50 - 1/5 X and 2) b - N/(M+N) X >>> 32 - 4/5 X

to find the rate at which compound C is formed is

The resulting reaction between the two chemicals is such that for each gram of A, 4 grams of B is used. It is observed that 30 grams of the compound C is formed in 10 minutes Determine the amount of C at time t if the rate of the reaction is proportional to the amounts of A and B remaining and if initially there are 50 grams of A and 32 grams of B. How much of the compound C is present at 15 minutes? Interpret the solution as t →∞ ?

In this example we will use the rate equation .

The rate equation: is the rate which a chemical A transforms into a second chemical B is proportional to the amount Q of A remaining untransformed at time t ,

DE= dQ/dt=KQ , K>0

Data:

A + 4B → C x(t):g X(0) = 0 , x(10)=30 , a=50g , b=32 , M=1 , N=4 , x(15)=?

Solution:

1) a - M/(M+N ) X >>> 50 - 1/5 X and 2) b - N/(M+N) X >>> 32 - 4/5 X

to find the rate at which compound C is formed is

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